Evaluate $\lim_{x \to 1} \dfrac{x^4 - \sqrt{x}}{\sqrt{x} - 1}$.

Evaluate limx → 1 dfracx⁴ - √(x)√(x) - 1. – Class 11 Mathematics

NCERT Exemplar · Q7

\lim_{x \to 1} \dfrac{x^4 - \sqrt{x}}{\sqrt{x} - 1}

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Substituting $u=\sqrt{x}$ turns the $0/0$ form into $\dfrac{u^8-u}{u-1}$ ; factoring gives $u\cdot\dfrac{u^7-1}{u-1}\to 1\cdot 7=7$ . The limit is $7$ .

Direct substitution gives $\dfrac{1-1}{1-1}=\dfrac{0}{0}$ , an indeterminate form.

1. Substitute $u=\sqrt{x}$ , so $x=u^2$ and $u\to 1$ :

\lim_{x\to 1}\frac{x^4-\sqrt{x}}{\sqrt{x}-1}=\lim_{u\to 1}\frac{u^8-u}{u-1}

2. Factor the numerator: $u^8-u=u(u^7-1)$ :

\lim_{u\to 1}\frac{u(u^7-1)}{u-1}=\lim_{u\to 1}u\cdot\frac{u^7-1}{u-1}

3. Use $\dfrac{u^7-1}{u-1}=u^6+u^5+u^4+u^3+u^2+u+1$ , which equals $7$ at $u=1$ :

$=1\cdot 7=7$ …

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Q.Evaluate $\lim_{x \to 1} \dfrac{x^4 - \sqrt{x}}{\sqrt{x} - 1}$ .

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